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Which of the following could be the value of G ?
[#permalink]
15 Dec 2021, 03:52
15 Dec 2021, 03:52
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Question Stats:
100% (01:20) correct
0% (00:00) wrong based on 2 sessions
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\(G^2\) < G
Which of the following could be the value of G ?
a) 1
b) \(23/7\)
c) \(7/23\)
d) -4
e) -2
I thought of solving it as
\(G^2\)-G<0
then G ( G-1)<0
G<0
G<1
So, the common numbers between the two is G<0
But this keeps us with two choices -2 & -4 . When we substitute the two numbers, they don't fit the equation.
So, if I substitute the the answers the only choice that fits the equation is 7/23
My confusion, how do we have a contradiction between the two concepts and the first is generally a rule of thumb in inequalities ?
Source: Math Strategies Manhattan
Which of the following could be the value of G ?
a) 1
b) \(23/7\)
c) \(7/23\)
d) -4
e) -2
I thought of solving it as
\(G^2\)-G<0
then G ( G-1)<0
G<0
G<1
So, the common numbers between the two is G<0
But this keeps us with two choices -2 & -4 . When we substitute the two numbers, they don't fit the equation.
So, if I substitute the the answers the only choice that fits the equation is 7/23
My confusion, how do we have a contradiction between the two concepts and the first is generally a rule of thumb in inequalities ?
Source: Math Strategies Manhattan
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Re: Which of the following could be the value of G ?
[#permalink]
15 Dec 2021, 06:39
15 Dec 2021, 06:39
1
Solving from starting
\(G^2\) < G
=> \(G^2\)-G<0
=> G (G-1)<0
Product of two values < 0 => these two values will have OPPOSITE Signs (Asmakan : You have taken both as negative which is not correct)
Now this will give us two cases
Case 1
G < 0 and G-1 > 0
=> G < 0 and G > 1
Intersection will give us NO SOLUTION in this case
Case 2
G > 0 and G-1 < 0
=> G > 0 and G < 1
=> 0 < G < 1
Only possible answer in this range is C
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Inequalities
\(G^2\) < G
=> \(G^2\)-G<0
=> G (G-1)<0
Product of two values < 0 => these two values will have OPPOSITE Signs (Asmakan : You have taken both as negative which is not correct)
Now this will give us two cases
Case 1
G < 0 and G-1 > 0
=> G < 0 and G > 1
Intersection will give us NO SOLUTION in this case
Case 2
G > 0 and G-1 < 0
=> G > 0 and G < 1
=> 0 < G < 1
Only possible answer in this range is C
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Inequalities
Re: Which of the following could be the value of G ?
[#permalink]
15 Dec 2021, 06:52
15 Dec 2021, 06:52
1
BrushMyQuant wrote:
Solving from starting
\(G^2\) < G
=> \(G^2\)-G<0
=> G (G-1)<0
Product of two values < 0 => these two values will have OPPOSITE Signs (Asmakan : You have taken both as negative which is not correct)
Now this will give us two cases
Case 1
G < 0 and G-1 > 0
=> G < 0 and G > 1
Intersection will give us NO SOLUTION in this case
Case 2
G > 0 and G-1 < 0
=> G > 0 and G < 1
=> 0 < G < 1
Only possible answer in this range is C
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Inequalities
\(G^2\) < G
=> \(G^2\)-G<0
=> G (G-1)<0
Product of two values < 0 => these two values will have OPPOSITE Signs (Asmakan : You have taken both as negative which is not correct)
Now this will give us two cases
Case 1
G < 0 and G-1 > 0
=> G < 0 and G > 1
Intersection will give us NO SOLUTION in this case
Case 2
G > 0 and G-1 < 0
=> G > 0 and G < 1
=> 0 < G < 1
Only possible answer in this range is C
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Inequalities
Based on your answer, here we don't treat the < as =.
For example, if we have [m]G^2=G[m]
we will solve it as
G(G-1)=0
Then G=0 or G=1
but because we have inequality here, then we have to consider it differently. Therefore, we have to see the signs and separate the two cases based on that, right ?
Secondly, do the solutions of the same variable have to intersect so we consider it as a solution ?
